Question

If $2{\tan }^2\theta ={\sec }^2\theta$ , then the general solution of $\theta$

• A. $n\pi +\frac{\pi }{4}$
• B. $n\pi -\frac{\pi }{4}$
• C. $n\pi \pm \frac{\pi }{4}$
• D. $2n\pi \pm \frac{\pi }{4}$

Answer 1

The correct option is C $n\pi \pm \frac{\pi }{4}$

$2{\tan }^2\theta ={\sec }^2\theta 2{\tan }^2\theta =1+{\tan }^2\theta {\tan }^2\theta =1\tan \theta =1\tan \theta =\tan \frac{\pi }{4}\theta =n\pi +\frac{\pi }{4}$